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Factor 32x^5 + 243y^5

  1. Mar 12, 2012 #1
    Completely factor 32x^5 + 243y^5

    okay i used synthetic divison and i referenced a factorign calculator to get the answer

    Answer:


    i used f(x) = (2x)^5 + (3y)^5

    then f(-3y/2)= (2x)^5 + (3y)^5 = 0 (factor thereom)

    using synthetic divison and factor (-3y/2)

    i got.. 32x^4 -48x^3y + 72x^2y^2 -108xy^3 + 162y^4



    The answer is (2x+3y)(16x^4 -24x^3y + 36x^2y^2 - 54xy^3 + 81y^4)



    i notice that the second peice (16x^4.......)

    the whole peice is actually half of what i long divided..


    so my problem is why am i not dividing to get that answer? and how can i put together what i knowÉ
     
  2. jcsd
  3. Mar 12, 2012 #2

    Mark44

    Staff: Mentor

    So what you have is
    (2x)5 + (3y)5 = (x - (-3y/2))(32x4 -48x3y + 72x2y2 -108xy3 + 162y4)
    = (x + 3y/2)(32x4 -48x3y + 72x2y2 -108xy3 + 162y4)

    If you pull a factor of 2 out of the long factor, and use it to multiply the first factor, you get what you show below.

     
  4. Mar 13, 2012 #3

    Just out of curiousity why am i allowed to divide a two out of the long factor, but i have to multiply it by the small factor? Like why is that allowed to happen, is there a rule of some sort... :O
     
  5. Mar 13, 2012 #4
    Maybe this will help clear the confusion up:

    Consider
    [itex]
    \begin{eqnarray*}
    (\frac{a}{2}+\frac{b}{2})(2a + 2b) &=& \frac{2}{2} (\frac{a}{2}+\frac{b}{2})(2a + 2b)\\
    &=& 2 \cdot (\frac{a}{2}+\frac{b}{2}) \cdot \frac{(2a + 2b)}{2}\\
    &=& (a+b)(a + b)
    \end{eqnarray*}
    [/itex]

    Does that help clear up what was done above?
     
  6. Mar 13, 2012 #5
    So is this kind of a choice, of expanding the 2 within the first factor and using the 1/2 and targeting the second factor peicce with it.. argh im just havin a rough time with this idea..

    the onyl connection i can make (whichi 'm not sure is true) is it looks like you multiply the equation by (2/2) to satisfy both the top and bottom of the equation, and then the rules of math just do the rest... but that's just a guess..

    Can u explain to me why you multiply the equation by 2/2
     
  7. Mar 13, 2012 #6
    I was just illustrating why you can divide one term by 2 and multiply the other by 2. It's the same thing as multiplying by [itex]\frac{2}{2} = 1[/itex]. I just figured it was easier to see in a smaller example.
     
  8. Mar 13, 2012 #7
    i understand that connection and looking at the smaller example in terms of the math(that is very clear).

    I think i was just thinking about it to hard, or perhaps not in the right manner. thanks a gain for your help man.
     
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